How to use the implicit function theorem in this case?

Question

Really hit a wall with this one:

Prove that the equations: $$2x+y+2z+u-v-1=0\\xy+z-u+2v-1=0\\yz+xz+u^2-v=0$$

define around $(u,v,x,y,z)=(1,1,-1,1,1)$ a single function $\phi(u,v)=(x(u,v),y(u,v),z(u,v))$. Find $D\phi(1,1)$

I believe I have to use the implicit function theorem, but I don't know how to apply it in this case.

Answer 1

basically what you have to do is set the function: $$ F(x,y,z,u,v)=\left(\begin{array}{c} F_1\\ F_2\\ F_3 \end{array}\right)=\left(\begin{array}{c} 2x+y+2z+u−v−1\\ xy+z−u+2v−1\\ yz+xz+u^2−v \end{array}\right) $$

Then you notice F(1,1,-1,1,1)=0. And you find the derivative with respect the variables you want to clear $$ \frac{\partial F}{\partial (x,y,z)}=\left(\begin{array}{c} \frac{\partial F_1}{\partial x} &\frac{\partial F_1}{\partial y} &\frac{\partial F_1}{\partial z} \\ \frac{\partial F_2}{\partial x} &\frac{\partial F_2}{\partial y} &\frac{\partial F_2}{\partial z} \\ \frac{\partial F_3}{\partial x} &\frac{\partial F_3}{\partial y} &\frac{\partial F_3}{\partial z} \end{array}\right)= \left(\begin{array}{c} 2 & 1 &2\\ y & x &1\\ z & z &y \end{array}\right) $$ And evaluated at the desired point $$ \frac{\partial F}{\partial (x,y,z)}(1,1,-1,1,1)=\left(\begin{array}{c} 2 & 1 &2\\ 1 & 1 &1\\ -1 & -1 &1 \end{array}\right) $$ And check that it has maximum rank, for instance as this is a square matrix we can check the determinant is non-zero. $$ det\left(\frac{\partial F}{\partial (x,y,z)}(1,1,-1,1,1)\right)=2\neq 0 $$ So by the implicit function theorem., there exist $\phi$ as you want. To get the derivative: $$ \frac{\partial \phi}{\partial(u,v)}(1,1)=-\left(\frac{\partial F}{\partial(x,y,z)}(1,1,-1,1,1)\right)^{-1}\frac{\partial F}{\partial(u,v)}(1,1,-1,1,1)\\ \frac{\partial \phi}{\partial(u,v)}(1,1)=\frac{1}{2}\left(\begin{array}{c} 2 & -2 &0\\ -3 & 4 &1\\ -1 & 0 &1 \end{array}\right) \left(\begin{array}{c} 1 & -1\\ -1 & 2\\ 2 &-1 \end{array}\right)\\ \frac{\partial \phi}{\partial(u,v)}(1,1)=\frac{1}{2}\left(\begin{array}{c} 4 & -6\\ -5 & 10\\ 1 & 0 \end{array}\right) $$

Answer 2

Consider $f: \Bbb R^2 \times \Bbb R^3 \to \Bbb R^3$,

$$f(u,v,x,y,z) =( 2x+y+2z+u-v-1, xy+z-u+2v-1, yz+xz+u^2-v)$$

We have:

$$\frac{\partial f}{\partial(x,y,z)} = \begin{pmatrix} 2 & 1 & 2 \\ y & x & 1 \\ z & z & x + y \end{pmatrix} $$

So:

$$\frac{\partial f}{\partial(x,y,z)}(1,1,-1,1,1) = \begin{pmatrix} 2 & 1 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & 0 \end{pmatrix}$$

which has determinant $3 \neq 0$ (hence invertible).

Finally, $f(1,1,-1,1,1) = 0$. So, by the implicit function theorem, there are two open sets $U \subset \Bbb R^2$ and $V \subset \Bbb R^3$, such that $(1,1) \in U$ and $(-1,1,1) \in V$, and a differentiable map $\phi: U \to V$, such that $f(u,v, \phi(u,v)) = 0$ for all $(u,v) \in U$. $\phi$ is the desired map. Now, by the second part of the IFT, we have:

$$D \Phi(1,1) = \left[ \frac{\partial f}{\partial(x,y,z)} (1,1,-1,1,1) \right]^{-1} \frac{\partial f}{\partial(u,v)}(1,1,-1,1,1)$$

Now:

$$\frac{\partial f}{\partial(u,v)} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \\ 2u & -1 \end{pmatrix}$$

At $(1,1,-1,1,1)$, this becomes:

$$\begin{pmatrix} 1 & -1 \\ -1 & 2 \\ 2 & -1 \end{pmatrix}$$

Thus;

$$D\phi(1,1) = \begin{pmatrix} 2 & 1 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 1 & -1 \\ -1 & 2 \\ 2 & -1 \end{pmatrix} = \frac13 \begin{pmatrix} -1 & 4 \\ 8 & -7 \\ -1 & -2 \end{pmatrix}$$